Question

Find the points on the y axis which is equidistant from the points (5 -2) and (-3,2)

Answer 1

Answer:

The point on y-axis is (0,-2)

Step-by-step explanation:

On y-axis means (0,y)

Let P(0,y) A(5,-2) B(-3,2)

According to question

PA=PB            [∵Equidistant]

Using distance formula

PA=$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_1)^2$

=$\sqrt{(5-0)^2+(-2-y)^2$

=$\sqrt{5^2+(-2)^2+y^2-2(-2)(y)$

=$\sqrt{25+4+y^2+4y$

=$\sqrt{y^2+4y+29}........Eq-1$

PB=$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

=$\sqrt{ (-3-0)^2+(2-y)^2$

=$\sqrt{(-3)^2+(2)^2+y^2-2(2)(y)$

=$\sqrt{9+4+y^2-4y$

=$\sqrt{y^2-4y+13}.........Eq-2$

PA=PB

$\sqrt{y^2+4y+29}$=$\sqrt{y^2-4y+13}$

Simplifying it

${y^2+4y+29}$=${y^2-4y+13}$

$y^2-y^2+4y+4y+29-13\\8y+16=0\\\\8y=-16\\y=-16/8\\y=-2$

∴ The point on y-axis is (0,-2)

Thank you